cassidy11
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Water is pouring into a bucket at a constant rate of 30cm^2 per second and is leaking out at a rate proportional to the volume of water already in the bucket.

1) show that at time t seconds the volume Vcm^2 of water in the bucket satisfies the differential equation dv/dt =30-kv where k is a positive constant

The bucket is initially empty
2) solve the differential equation and show that V=A +Be^-kt giving the values of A and B in terms of k.
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RDKGames
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(Original post by cassidy11)
Water is pouring into a bucket at a constant rate of 30cm^2 per second and is leaking out at a rate proportional to the volume of water already in the bucket.

1) show that at time t seconds the volume Vcm^2 of water in the bucket satisfies the differential equation dv/dt =30-kv where k is a positive constant

The bucket is initially empty
2) solve the differential equation and show that V=A +Be^-kt giving the values of A and B in terms of k.
What have you tried? Also your volume dimensions are slightly off...
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cassidy11
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(Original post by RDKGames)
What have you tried?
I honestly don't know where to start i set t as 0 but didn't know where to go then
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cassidy11
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(Original post by RDKGames)
What have you tried? Also your volume dimensions are slightly off...
I know I was just thinking that lol but that's what the question says
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RDKGames
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(Original post by cassidy11)
I honestly don't know where to start i set t as 0 but didn't know where to go then
Okay so formulate the information given.

\frac{dV}{dt} is your change of volume

You have 30\text{cm}^3\text{s}^{-1} going in

Then you have some going out, denote this as O\text{cm}^3\text{s}^{-1} and we know O \propto V \Rightarrow O=kV where k>0

So your change of volume at each instant of time is defined by how much is going in take away the amount that is going out.

\frac{dV}{dt}=30-O = 30-kV
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cassidy11
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(Original post by RDKGames)
Okay so formulate the information given.

\frac{dV}{dt} is your change of volume

You have 30\text{cm}^3\text{s}^{-1} going in

Then you have some going out, denote this as O\text{cm}^3\text{s}^{-1} and we know O \propto V \Rightarrow O=kV where k>0

So your change of volume at each instant of time is defined by how much is going in take away the amount that is going out.

\frac{dV}{dt}=30-O = 30-kV
Thank you very much do you know how to solve it?
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Zacken
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(Original post by cassidy11)
Thank you very much do you know how to solve it?
It's a seperable first order linear differential equation: you should know how to solve it.
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